Theorem 1

Let $x_1, \ldots, x,$ be elements of the ring of $v'$ whose canonical images $\overline{x}_i$ in $k'$ are algebraically independent over $k$ and $y_1, \ldots, y_r$ elements of $K'$ such that the canonical images of the $v'(y_j)$ in $\Gamma'/\Gamma$ are linearly independent over $\mathbf{Z}$. Then the $r+s$ elements $x_1, \ldots, x, y_1, \ldots, y_r$ of $K'$ are algebraically independent over $K$; the restriction of $v'$ to $K(x_1, \ldots, x, y_1, \ldots, y_r)$ admits $k(\overline{x}_1, \ldots, \overline{x}_s)$ as residue field and

as order group.

Our assertion is obvious if $r+s = 0$. We argue by induction on $r+s$. If $r' \leq r, s' \leq s$ and $r'+s' < r+s$, the induction hypothesis shows that the hypotheses of Theorem 1 hold if $K$ is replaced by $K(x_1, \ldots, x_s, y_1, \ldots, y_{r'})$ and the families $(x_1, \ldots, x_s), (y_1, \ldots, y_r)$ by $(x_{s'+1}, \ldots, x_s), (y_{r'+1}, \ldots, y_r)$. The problem is therefore reduced to one of the two following cases:

(a) There is an element $x$ in the ring $v'$ such that $\overline{x}$ is transcendental over $k$; then it is necessary to show that $x$ is transcendental over $K$ and that the restriction of $v'$ to $K(x)$ admits $k(\overline{x})$ as residue field and $\Gamma$ as order group.

(b) There is an element $y$ in $K'$ such that the relations $nv'(y) \in \Gamma$ and $n \in \mathbf{Z}$ imply $n = 0$; it is necessary to show that $y$ is transcendental over $K$ and that the restriction of $v'$ to $K(y)$ admits $k$ as residue field and $\Gamma + \mathbf{Z}v'(y)$ as order group.

Now Proposition 1 of § 8, no. 1 shows that $x$ (resp. $y$) cannot be algebraic over $K$. The other assertions of (a) (resp. (b)) follow immediately by Proposition 2 (resp. Proposition 1) of no. 1.